Question

A 0.40-kg iron horseshoe, just forged and very hot is dropped into 1.25 L of water in a 0.28-kg iron pot initially at 20.0?C. The value of specific heat for iron is 450 J/kg?C? , and for water is 4186 J/kg?C? If the final equilibrium temperature is 25.0 ?C, Determine the initial temperature of the hot horseshoe.

Answer 1

given
m_horseshoe = 0.4 kg
m_water = 1.25 kg
m_Iron = 0.28 kg

let T is the initial tempearture of iron horseshoe

use,

Heat lost by iron horseshoe = Hwat gained by water + Iron pot

m_horseshoe*C_iron*(T - 25) = m_water*C_water*(25 - 10) + m_Iron*C_Iron*(25 - 20)

0.4*450*(T - 25) = 1.25*4186*5 + 0.28*450*5

0.4*450*T - 0.4*450*25 = 1.25*4186*5 + 0.28*450*5

T = (0.4*450*25 + 1.25*4186*5 + 0.28*450*5)/(0.4*450)

= 174 degrees celsius <<<<<<<------Answer